Constructing an infinite chain of subfields of 'hyper' algebraic numbers?

This has now been cross posted to MO.

Let $$F$$ be a subset of $$\mathbb{R}$$ and let $$S_F$$ denote the set of values which satisfy some generalized polynomial whose exponents and coefficients are drawn from $$F$$. That is, we let $$S_F$$ denote $$\bigg \{x \in \mathbb{R}: 0=\sum_{i=1}^n{a_i x^{e_i}}: e_i \in F \text{ distinct}, a_i\in F \text{ non-zero}, n\in \mathbb{N} \bigg \}$$

Then $$S_{\mathbb{\mathbb{Q}}}$$ is the set of algebraic real numbers and we start to see the beginnings of a chain:

$$\mathbb{Q} \subsetneq S_\mathbb{Q} \subsetneq S_{S_\mathbb{Q}}$$

Main Question

Does this chain continue forever? That is, we let $$A_0= \mathbb{Q}$$ and let $$A_{n+1}=S_{A_{n}}$$. Is it the case that $$A_n \subsetneq A_{n+1}$$ for all $$n\in\mathbb{N}$$?

Other curiosities:

Is $$A_i$$ always a field? Perhaps, the argument is analogous to this. Or maybe this is just the case in a more general setting: Is it the case that $$F \subset \mathbb{R}$$, a field implies that $$S_F$$ is a field?

Is it possible to see that $$e\notin \cup A_i$$? Perhaps this is just a tweaking of LW Theorem.

• Every set in the chain is countable, so you certainly don't hit $\mathbb{R}$ at any point. – Patrick Stevens Nov 26 '18 at 19:19
• How do you propose to define $x^\alpha$ if $\alpha \notin \mathbb{Q}$? – Hans Engler Nov 26 '18 at 19:24
• I'm still somewhat concerned about the definition of $x^\alpha$ when $x<0$ and $\alpha$ is not an integer. But, there is a cool question in here (possibly a very difficult one). – Jyrki Lahtonen Dec 3 '18 at 4:31
• Mason, I had some simple worries in my mind. Like is $S_F$ necessarily closed under negation? – Jyrki Lahtonen Dec 3 '18 at 14:06
• I don’t think we can rule out $e\in A_3$. My impression is that we know almost nothing about the transcendence of numbers like $3^\sqrt6-4^\sqrt3-1$. – user210229 Dec 6 '18 at 22:35